Difference Paradox

Consider two natural numbers $n_1$ and $n_2$, out of which one is twice as large as the other. We are not told whether $n_1$ is larger or $n_2$, we can state following two propositions: PROPOSITION 1: The difference $n_1-n_2$, if $n_1 >n_2$, is different from the difference $n_2-n_1$, if $n_2 >n_1$. PROPOSITION 2: The difference $n_1-n_2$, if $n_1 >n_2$, is the same as the difference $n_2-n_1$, if $n_2 >n_1$. Moving on the proofs: PROOF OF PROPOSITION 1: Let $n_1 > n_2$, then $n_1=2n_2$.…

Smart Fallacies: i=1, 1= 2 and 1= 3

This mathematical fallacy is due to a simple assumption, that $ -1=\dfrac{-1}{1}=\dfrac{1}{-1}$ . Proceeding with $ \dfrac{-1}{1}=\dfrac{1}{-1}$ and taking square-roots of both sides, we get: $ \dfrac{\sqrt{-1}}{\sqrt{1}}=\dfrac{\sqrt{1}}{\sqrt{-1}}$ Now, as the Euler's constant $ i= \sqrt{-1}$ and $ \sqrt{1}=1$ , we can have $ \dfrac{i}{1}=\dfrac{1}{i} \ldots \{1 \}$ $ \Rightarrow i^2=1 \ldots \{2 \}$ . This is complete contradiction to the fact that $ i^2=-1$ . Again, as $ \dfrac{i}{1}=\dfrac{1}{i}$ or, $ i^2=1$ or, $ i^2+2=1+2$ or, $ -1+2=3$ $ 1=3…